course Phy 121 ???€???zK??????????assignment #021021. projectiles 2
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13:16:49 `q001. Note that this assignment contains 3 questions. . A projectile has an initial velocity of 12 meters/second, entirely in the horizontal direction. After falling to a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's velocity vector?
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RESPONSE --> thru Newton's Laws i figured the 'dt to be .78sec. that allows me to draw the vector on the map. one side of the new right triangle will be .78, the other side 3, therefore the mag of the vector, which would be equal to the hypotenuese, is 3.1 m/s. the direction is 285 degrees. confidence assessment: 3
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13:17:50 To answer this question we must first determine the horizontal and vertical velocities of the projectile at the instant it first encounters the floor. The horizontal velocity will remain at 12 meters/second. The vertical velocity will be the velocity attained by a falling object which is released from rest and allowed to fall three meters under the influence of gravity. Thus the vertical motion will be characterized by initial velocity v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8 meters/second ^ 2. The fourth equation of motion, vf^2 = v0^2 + 2 a `ds, yields final vel in y direction: vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second. Since we took the acceleration to be in the positive direction the final velocity will be + 7.7 meters/second. This final velocity is in the downward direction. On a standard x-y coordinate system, this velocity will be directed along the negative y axis and the final velocity will have y coordinate -7.7 m/s and x coordinate 12 meters/second. The magnitude of the final velocity is therefore `sqrt((12 meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second, approximately. The direction of the final velocity will therefore be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees, very approximately, as measured in the counterclockwise direction from the positive x axis. The direction of the projectile at this instant is therefore 35 degrees below horizontal. This angle is more commonly expressed as 360 degrees + (-35 degrees) = 325 degrees.
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RESPONSE --> my mistake was putting the x coordinate at the 'dt which was .78, if i had done it correctly using 12 m/s as the x coordinate i would have been fine here. self critique assessment: 3
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13:33:15 `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
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RESPONSE --> i think i could get this one except that i don't know how to get the initial velocity of the object's vertical motion, it has to be greater than 0 i know the mag of the vector is 18.5 m/s but i don't know how to utilize that info, if i even need it. confidence assessment: 0
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13:34:32 To determine the time required to reach the level surface we need only analyze the vertical motion of the projectile. The acceleration in the vertical direction will be 9.8 meters/second ^ 2 in the downward direction, and the displacement will be 12 meters in the downward direction. Taking the initial velocity to be upward into the right, we situate our x-y coordinate system with the y direction vertically upward and the x direction toward the right. Thus the initial velocity in the vertical direction will be equal to the y component of the initial velocity, which is v0y = 20 meters/second * sine (30 degrees) = 10 meters/second. Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8 meters/second ^ 2, we see that we can find the time `dt required to reach the level surface using either the third equation of motion `ds = v0 `dt + .5 a `dt^2, or we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find vf after which we can easily find `dt. To avoid having to solve a quadratic in `dt we choose to start with the fourth equation. We obtain vf = +-`sqrt ( (10 meters/second) ^ 2 + 2 * (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second, approximately. Since we know that the final velocity will be in the downward direction, we choose vf = -18.3 meters/second. We can now find the average velocity in the y direction. Averaging the initial 10 meters/second with the final -18.3 meters/second, we see that the average vertical velocity is -4.2 meters/second. Thus the time required for the -12 meters displacement is `dt = `ds / vAve = -12 meters/(-4.2 meters/second) = 2.7 seconds.
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RESPONSE --> i see how to get the initial velocity. after that i believe i've got it. self critique assessment: 3
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13:37:15 `q003. What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes the flat surface.
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RESPONSE --> i know the v0 to be 20m/s and the length it travels before leaving the platform or whatever is 18.5 m/s so it must take .93 sec plus the 2.7 sec it took after it landed making the total from beginning to end 3.63 sec or so confidence assessment: 2
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13:38:54 The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing that the projectile travels for 2.7 seconds we can easily find the horizontal range. The horizontal velocity of the projectile is simply the x component of the velocity: horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second. Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters, approximately.
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RESPONSE --> woops, i answered a question i must have made up in my mind. sorry. this exercise did cause me to think in ways i haven't before regarding projectiles. i am glad i learned about the vertical velocity of the projectiles traveling up and horizontally. i'm sure i'll need that for later. self critique assessment: 3
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????????D?j???assignment #021 g???z???????Physics I 07-15-2007
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18:19:49 Explain how to obtain the final speed and direction of motion of a projectile which starts with known velocity in the horizontal direction and falls a known vertical distance, using the analysis of vertical and horizontal motion and vectors.
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RESPONSE --> if you know the vertical 'ds, you can figure the 'dt because you know the acceleration is 9.8 m/s^2 the speed, or velocity, can be the x coordinate vector and by this time we will have figured the y coordinate vector which is its vertical speed we find the direction by finding the arctan of y/x
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18:20:14 ** The horizontal velocity is unchanging so the horizontal component is always equal to the known initial horizontal velocity. The vertical velocity starts at 0, with acceleration thru a known distance at 9.8 m/s^2 downward. The final vertical velocity is easily found using the fourth equation of motion. We therefore know the x (horizontal) and y (vertical) components of the velocity. Using the Pythagorean Theorem and arctan (vy / vx) we find the speed and direction of the motion. **
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RESPONSE --> i got this i believe
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18:24:32 Give at least three examples of vector quantities for which we might wish to find the components from magnitude and direction. Explain the meaning of the magnitude and the direction of each, and explain the meaning of the vector components.
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RESPONSE --> 'ds, acceleration, and force are all vector quantities that we could find the components from their magnitude and direction we would find the magnitude of a force vector, for example of x component 20N and y component 15N by taking the square root of the sum of their squares which in this case would be 625, we would find the direction by finding the arctan of y/x which here would be 36.8 degrees
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